# How to show $K_p = \frac{\det II}{\det I}$?

where $$K_p$$ is the gaussian curvature, $$I$$ is the first fundamental form and $$II$$ is the second fundamental form.

Suppose $$\xi(u,v)$$ is my surface patch. I then computed $$II$$ as $$II_{11} = \langle n_p, \frac{\partial^2 \xi}{ \partial u^2} \rangle$$, $$II_{12} = II_{21} = \langle n_p, \frac{\partial^2 \xi}{ \partial u \partial v} \rangle$$, $$II_{22} = \langle n_p, \frac{\partial^2 \xi}{ \partial v^2} \rangle$$.

Then, $$\det II = \langle n_p, \frac{\partial^2 \xi}{ \partial u^2} \rangle \cdot \langle n_p, \frac{\partial^2 \xi}{ \partial v^2} \rangle - \left(\langle n_p, \frac{\partial^2 \xi}{ \partial u \partial v} \rangle \right)^2$$

And I know $$\det I = \left|\left| \frac{\partial \xi}{\partial u} \times \frac{\partial \xi}{\partial v} \right|\right|^2$$

But from here I don't know how to proceed.

Let $$\zeta(u,v)$$ your surface patch.The gaussian curvature of $$\zeta$$ is $$K=k_1k_2,$$ where $$k_1$$ and $$k_2$$ are the principal curvatures. The first fundamental form can be expressed as $$Edu^2+2Fdudv+G^2dv^2,$$ and the second fundamental form as $$Ldu^2+2Mdudv+Ndv^2.$$ Let $$$$A=\begin{pmatrix}E&F\\ F& G\end{pmatrix},$$$$ and $$$$B=\begin{pmatrix}L&M\\ M&N\end{pmatrix}.$$$$ By definition $$k_1$$ and $$k_2$$ are the eigenvalues of $$A^{-1}B$$, so $$K=k_1k_2=\big|A^{-1}B\big |=\frac{LN-M^2}{EG-F^2}$$